Wednesday 13 June 2012 Morning

Wednesday 13 June 2012 Morning A2 GCE CHEMISTRY A F325 Equilibria, Energetics and Elements *F314750612* Candidates answer on the Question Paper. OCR supplied materials: Data Sheet for Chemistry A (inserted) Other materials required: Scientific calculator Duration: 2 hours * F 3 2 5 * INSTRUCTIONS TO CANDIDATES The insert will be found in the centre of this document. Write your name, centre number and candidate number in the boxes above. Please write clearly and in capital letters. Use black ink. Pencil may only be used for graphs and diagrams where they appear. Read each question carefully. Make sure that you know what you have to do before starting your answer. Write your answer to each question in the space provided. If additional space is required, you should use the lined pages at the end of this booklet. The question number(s) must be clearly shown. Answer all the questions. Do not write in the bar codes. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. Where you see this icon you will be awarded marks for the quality of written communication in your answer. This means for example you should: ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear; organise information clearly and coherently, using specialist vocabulary when appropriate. You may use a scientific calculator. A copy of the Data Sheet for Chemistry A is provided as an insert with this question paper. You are advised to show all the steps in any calculations. The total number of marks for this paper is 100. This document consists of 20 pages. Any blank pages are indicated. OCR 2012 [T/500/7837] DC (SHW (CGW) 00660 12/10) 43713/4 OCR is an exempt Charity Turn over

2 Answer all the questions. 1 Lattice enthalpy can be used as a measure of ionic bond strength. Lattice enthalpies can be determined indirectly using Born Haber cycles. The table below shows the enthalpy changes that are needed to determine the lattice enthalpy of lithium fluoride, LiF. enthalpy change energy / kj mol 1 1st electron affinity of fluorine 328 1st ionisation energy of lithium +520 atomisation of fluorine +79 atomisation of lithium +159 formation of lithium fluoride 616 (a) Define the term lattice enthalpy............. [2] (b) The diagram below shows an incomplete Born Haber cycle that would allow the lattice enthalpy of lithium fluoride to be determined. (i) On the four dotted lines, add the species present, including state symbols....... Li + (g) + F (g)...... OCR 2012 LiF(s) [4]

3 (ii) Calculate the lattice enthalpy of lithium fluoride. lattice enthalpy =... kj mol 1 [2] (c) The change that produces lattice enthalpy is spontaneous but has a negative entropy change. Why is this change able to take place spontaneously?......... [1] (d) The lattice enthalpies of sodium fluoride, sodium chloride and magnesium fluoride are shown below. compound lattice enthalpy / kj mol 1 sodium fluoride 918 sodium chloride 780 magnesium fluoride 2957 Explain the differences between these lattice enthalpies. In your answer, your explanation should show how different factors affect lattice enthalpy............................ [3] OCR 2012 [Total: 12] Turn over

2 A chemist carries out an investigation on the equilibrium system shown below. 4 2CO(g) + 2NO(g) 2CO 2 (g) + N 2 (g) ΔH = 788 kj mol 1 The chemist mixes 0.46 mol of CO with 0.45 mol of NO. The mixture is left to reach equilibrium at constant temperature. The student analyses the equilibrium mixture and finds that 0.25 mol NO remains. The total volume of the equilibrium mixture is 1.0 dm 3. (a) (i) Write the K c expression for this equilibrium. [1] (ii) What are the units of this equilibrium constant?... [1] (iii) Determine the value of K c for this equilibrium mixture. Show all your working. K c =... [4] OCR 2012

(iv) 5 What does your value of K c suggest about the position of equilibrium in this experiment?... [1] (b) The chemist increases both the temperature and the pressure of the equilibrium mixture. The mixture is left to reach equilibrium again. (i) What is the effect, if any, on the value of K c? Explain your answer.... [1] (ii) Explain why it is difficult to predict what would happen to the position of equilibrium after these changes in temperature and pressure.... [2] [Total: 10] OCR 2012 Turn over

6 3 Butanoic acid, CH 3 (CH 2 ) 2 COOH, is the butter acid, formed when butter turns rancid and tastes sour. A student prepares an aqueous solution of butanoic acid with a concentration of 0.250 mol dm 3. The K a of butanoic acid is 1.51 10 5 mol dm 3. (a) (i) Write the expression for the acid dissociation constant of butanoic acid. [1] (ii) Calculate the pk a of butanoic acid. pk a =... [1] (iii) Calculate the ph of the 0.250 mol dm 3 butanoic acid. Give your answer to two decimal places. (b) The student adds aqueous butanoic acid to magnesium. ph =... [3] The student then adds aqueous butanoic acid to aqueous sodium carbonate. (i) Write the ionic equation for the reaction between aqueous butanoic acid and magnesium.... [1] (ii) Write the ionic equation for the reaction between aqueous butanoic acid and aqueous sodium carbonate.... [1] OCR 2012

7 (c) The student adds 50.0 cm 3 of 0.250 mol dm 3 butanoic acid to 50.0 cm 3 of 0.0500 mol dm 3 sodium hydroxide. A buffer solution forms. (i) Explain why a buffer solution forms.... [2] (ii) Calculate the ph of the buffer solution. The K a of butanoic acid is 1.51 10 5 mol dm 3. Give your answer to two decimal places. ph =... [5] (d) The student adds methanoic acid, HCOOH (K a = 1.82 10 4 mol dm 3 ), to butanoic acid. A reaction takes place to form an equilibrium mixture containing two acid base pairs. Complete the equilibrium below and label the conjugate acid base pairs. HCOOH + CH 3 (CH 2 ) 2 COOH... +............... [2] [Total: 16] OCR 2012 Turn over

4 Nitrogen dioxide reacts with ozone as shown below. 8 2NO 2 (g) + O 3 (g) N 2 O 5 (g) + O 2 (g) (a) The kinetics of the reaction between NO 2 and O 3 was investigated and the following experimental results were obtained. experiment [NO 2 (g)] [O 3 (g)] initial rate / mol dm 3 / mol dm 3 / mol dm 3 s 1 1 0.00150 0.00250 4.80 10 8 2 0.00225 0.00250 7.20 10 8 3 0.00225 0.00500 1.44 10 7 (i) Determine the rate equation and rate constant for the reaction of NO 2 (g) and O 3 (g). In your answer you should make clear how your conclusions fit with the experimental results.... [8] OCR 2012

(ii) Suggest a possible two-step mechanism for this reaction. 9... [2] (b) The feasibility of the reaction between NO 2 and O 3 is influenced by the enthalpy change and entropy change of the reaction and the temperature. 2NO 2 (g) + O 3 (g) N 2 O 5 (g) + O 2 (g) ΔH = 198 kj mol 1 ΔS = 168 J K 1 mol 1 (i) Explain why this reaction has a negative entropy change.... [2] (ii) Calculate the value of ΔG, in kj mol 1, at 25 C for the reaction of NO 2 with O 3. ΔG =... kj mol 1 [3] (iii) State and explain how the feasibility of this reaction will change with increasing temperature.... [2] [Total: 17] OCR 2012 Turn over

5 Iron, copper and platinum are examples of transition elements. (a) Define the term transition element. 10 Show that iron fits this definition by use of full electron configurations of iron as the element and in its common oxidation states......................... [4] (b) Describe one precipitation reaction and one ligand substitution reaction of copper in the +2 oxidation state. Your answer should include reagents, relevant observations and balanced equations........................................ [6] OCR 2012

11 (c) Platinum is an extremely unreactive transition element. However, platinum does take part in a redox reaction with aqua regia, a mixture of concentrated hydrochloric and nitric acids. Two products of this reaction are hexachloroplatinic acid, H 2 PtCl 6, and nitrogen dioxide, NO 2. (i) Use oxidation states to show that this is a redox reaction.... [2] (ii) Write an equation for the reaction of platinum metal with aqua regia.... [2] (d) Ammonium hexachloroplatinate, (NH 4 ) 2 PtCl 6, is a complex of platinum used in platinum plating. Ammonium hexachloroplatinate contains the hexachloroplatinate ion. Draw a 3-D diagram to show the shape of a hexachloroplatinate ion. On your diagram, show the charge on the ion the value of the bond angle. [3] OCR 2012 Turn over

(e) Oxaliplatin is a neutral complex of platinum(ii) used in cancer treatment. 12 A molecule of oxaliplatin has a square planar shape about the metal ion with two bidentate ligands. The structure of oxaliplatin is shown below. H 2 N O O Pt N H 2 O O (i) What is meant by a bidentate ligand?... [2] (ii) In the boxes below, show the structures of the two bidentate ligands in oxaliplatin. [2] [Total: 21] OCR 2012

13 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE TURN OVER FOR QUESTION 6 OCR 2012 Turn over

6 Standard electrode potentials for eight redox systems are shown in Table 6.1. 14 You will need to use this information throughout this question. redox system half-equation E o / V 1 2H + (aq) + 2e H 2 (g) 0.00 2 Fe 3+ (aq) + e Fe 2+ (aq) +0.77 3 Cr 2 O 2 7 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) +1.33 4 O 2 (g) + 4H + (aq) + 4e 2H 2 O(l) +1.23 5 Cu 2+ (aq) + 2e Cu(s) +0.34 6 CO 2 (g) + 2H + (aq) + 2e HCOOH(aq) 0.22 7 HCOOH(aq) + 2H + (aq) + 2e HCHO(aq) + H 2 O(l) +0.06 8 Cr 3+ (aq) + 3e Cr(s) 0.74 Table 6.1 (a) A student sets up a standard cell in the laboratory based on redox systems 2 and 8. His circuit allows him to measure the standard cell potential. (i) Draw a labelled diagram to show how the student could have set up this cell to measure its standard cell potential. [3] (ii) Write down the overall cell reaction.... [1] (iii) Write down the standard cell potential. standard cell potential =... V [1] (b) Select from Table 6.1, the strongest oxidising agent.... [1] OCR 2012

15 (c) Using the redox systems in Table 6.1, construct an equation for a reaction between acidified dichromate(vi) ions and methanoic acid, HCOOH. Rather than using [O] or [H], your equation must show the actual reactants and products. (d) A student added some chromium metal to an acidified solution containing copper(ii) ions. A reaction took place. The student concluded that chromium is more reactive than copper. [2] (i) Explain, in terms of their electrode potentials, why chromium is more reactive than copper in this reaction.... [2] (ii) When this experiment was carried out, the student observed some bubbles of a gas. Suggest an explanation for this observation.... [1] (e) Methanoic acid, HCOOH, has the common name of formic acid. Direct-Formic Acid Fuel Cells (DFAFCs) are being developed for use in small, portable electronics such as phones and laptop computers. In this fuel cell, methanoic acid (the fuel) reacts with oxygen to generate a cell potential. (i) Predict the standard cell potential of a DFAFC. standard cell potential =... V [1] (ii) Suggest two advantages of using methanoic acid as the fuel in a fuel cell rather than hydrogen.... [2] [Total: 14] OCR 2012 Turn over

16 7 Potassium manganate(vii) can be prepared in the laboratory by a two-step synthesis starting from manganese(iv) oxide. Step 1 Step 2 In this step, manganese(iv) oxide is heated strongly with potassium hydroxide and potassium chlorate(v), a powerful oxidising agent. Manganese(IV) oxide, MnO 2, is oxidised to manganate(vi) ions. Potassium manganate(vi) is separated from the alkaline mixture from step 1 as a green solid. In this step, potassium manganate(vi) is heated in water. Manganate(VI) ions disproportionate forming manganate(vii) ions and a precipitate of manganese(iv) oxide. (a) In step 1, a redox reaction takes place. Add the correct number of electrons to the correct sides of the incomplete oxidation and reduction half-equations shown below. MnO 2 + 4OH MnO 4 2 + 2H 2 O 3H 2 O + Cl O 3 6OH + Cl [2] (b) In step 2, an equilibrium is set up. 3MnO 4 2 (aq) + 2H 2 O(l) 2MnO 4 (aq) + MnO 2 (s) + 4OH (aq) The equilibrium position can be shifted by bubbling carbon dioxide gas through the mixture. Suggest, with the aid of an equation, how the equilibrium position shifts................... [3] OCR 2012

17 (c) Aqueous potassium manganate(vii), KMnO 4, in acidic conditions can be used in analysis. A student analyses a sample of sodium sulfite, Na 2 SO 3, using the following method. The student dissolves 0.720 g of impure sodium sulfite in water. The solution is made up to 100.0 cm 3. The student titrates 25.0 cm 3 of this solution with 0.0200 mol dm 3 KMnO 4 under acidic conditions. The volume of KMnO 4 (aq) required to reach the end-point is 26.2 cm 3. The equation for the reaction is shown below. 2MnO 4 + 6H + + 5SO 3 2 2Mn 2+ + 5SO 4 2 + 3H 2 O Determine the percentage purity of the sample of sodium sulfite. percentage purity =... % [5] [Total: 10] END OF QUESTION PAPER OCR 2012

18 ADDITIONAL PAGE If additional space is required, you should use the lined pages below. The question number(s) must be clearly shown. OCR 2012

19 ADDITIONAL PAGE OCR 2012

20 ADDITIONAL PAGE Copyright Information OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE. OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. OCR 2012

GCE Chemistry A Advanced GCE Unit F325: Equilibria, Energetics and Elements Mark Scheme for June 2012 Oxford Cambridge and RSA Examinations

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, OCR Nationals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. OCR 2012 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: 0870 770 6622 Facsimile: 01223 552610 E-mail: publications@ocr.org.uk

Annotations available in Scoris. Annotation Meaning Benefit of doubt given Contradiction Incorrect response Error carried forward Ignore Not answered question Benefit of doubt not given Power of 10 error Omission mark Rounding error Error in number of significant figures Correct response 1

Abbreviations, annotations and conventions used in the detailed Mark Scheme (to include abbreviations and subject-specific conventions). Annotation DO NOT ALLOW IGNORE ALLOW Meaning Answers which are not worthy of credit Statements which are irrelevant Answers that can be accepted ( ) Words which are not essential to gain credit ECF AW ORA Underlined words must be present in answer to score a mark Error carried forward Alternative wording Or reverse argument The following questions should be annotated with ticks, crosses, etc. Annotations should be placed to clearly show where they apply within the body of the text (i.e. not in margins) Question 1(b)(i), (c), (d); Question 2(a)(iii); Question 3c(ii); Question 4a(i), (b)(iii); Question 5(b); Question 7(b), (c). 2

Question Answer Marks Guidance 1 (a) (The enthalpy change that accompanies) the formation of one mole of a(n ionic) compound IGNORE 'Energy needed' OR energy required from its gaseous ions (under standard conditions) 2 ALLOW as alternative for compound: lattice, crystal, substance, solid, product Note: 1st mark requires 1 mole 2nd mark requires gaseous ions IF candidate response has 1 mole of gaseous ions, award 2nd mark but NOT 1st mark IGNORE reference to constituent elements IGNORE: Li + (g) + F (g) LiF(s) Question asks for a definition, not an equation 3

Question Answer Marks Guidance 1 (b) (i) 1. Mark Line 1 first as below (right or wrong) 2. Mark Line 4 as below (right or wrong) ANNOTATIONS MUST BE USED 3. Mark difference in species on Line 1 and Line 2 MUST match one of the enthalpy changes in the table: atomisation of Li(s) atomisation of ½F 2 (g) first ionisation energy of Li(g) --------------------------------------------------- ALLOW marks by ECF as follows: Follow order at top of Answer column 4. Repeat for differences on Line 2 and Line 3 4 Li + (g) + F(g) + e ALLOW atomisation of ½F 2 (g) before atomisation of Li(s): ALLOW ionisation of Li(g) before atomisation of ½F 2 (g): 3 2 1 Li(g) + F(g) Li(g) + 1 / 2 F 2 (g) Li(s) + 1 / 2 F 2 (g) Correct species and state symbols required for all marks IF an electron has formed, it MUST be shown as e OR e 4 4 3 2 1 Li + (g) + F(g) + e Li(g) + F(g) Li(s) + 1 / 2 F 2 (g) Li(s) + F(g) Li(g) Li + (g) + F(g) + e e required for marks involving Line 3 AND Line 4 Common errors Line 4: Missing e and rest correct 3 marks Line 1: IF ½F 2 (g) is NOT shown 2 max [Line 4 and Li(s) Li(g) ] e.g., for F(g), F(s), F(l), F(aq), F 2 (g) DO NOT ALLOW Fl when first seen but credit subsequently 4 3 2 1 Li + (g) + e + 1 / 2 F 2 (g) + 1 / 2 F 2 (g) Li(s) + 1 / 2 F 2 (g) 4

Question Answer Marks Guidance 1 (b) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 1046 (kj mol 1 ) award 2 marks -------------------------------------------------------------------- ( 616) = (+159) + (+79) + (+520) + ( 328) + H LE (LiF) IF there is an alternative answer, check the list below for marking of answers from common errors ------------------------------------------------------------- ALLOW for 1 mark: OR +1046 wrong sign H LE (LiF) = ( 616) [ (+159) + (+79) + (+520) + ( 328) ] 186 +186 +430 instead of 430 +616 instead of 616 1006.5 (+79) H at (F) halved to +39.5 = 616 430 1702 wrong sign for 328 = 1046 (kj mol 1 ) 2 Any other number: CHECK for ECF from 1st marking point for expressions with ONE error only e.g. one transcription error: e.g. +195 instead of +159 (c) ANNOTATIONS MUST BE USED H < T S OR H T S < 0 OR H is more negative than T S OR Negative value of H is more significant than negative value of T S NOTE IGNORE comments about G 1 ALLOW exothermic for negative ALLOW a negative lattice energy value ALLOW H is negative AND magnitude of H > magnitude of T S IGNORE ONLY magnitude of H > magnitude of T S 5

Question Answer Marks Guidance 1 (d) For FIRST TWO marking points, assume that the following refer to ions, Mg 2+, etc. For ions, ALLOW atoms For Mg 2+, Na +, Cl and F, ALLOW symbols: Mg, Na, Cl and F ALLOW names: magnesium, sodium, chlorine, chloride, fluorine, fluoride i.e. ALLOW Mg has a smaller (atomic) radius DO NOT ALLOW molecules ALLOW Fl for F For THIRD marking point, IONS must be used Comparison of size of anions Chloride ion OR Cl is larger (than F ) OR Cl has smaller charge density (than F ) Comparison of size AND charge of cations Mg 2+ is smaller (than Na + ) AND Mg 2+ has a greater charge (than Na + ) Comparison of attraction between ions F has greater attraction for Na + / + ions AND Mg 2+ has greater attraction for F / ions Quality of Written Communication: --------------------------------------------------------------------------- Third mark needs to link ionic size and ionic charge with the attraction that results in lattice enthalpy 3 Total 12 ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- ORA F is smaller OR F has a larger charge density IGNORE just Cl is large comparison required ORA: Na + is larger AND Na + has a smaller charge IGNORE just Mg 2+ is small comparison required ALLOW greater charge density for greater charge but NOT for smaller size + AND IONS must be used for this mark IGNORE greater attraction between ions in NaF AND MgF 2 + AND ions OR oppositely charged ions are required ASSUME attraction to be electrostatic unless stated otherwise: e.g. DO NOT ALLOW nuclear attraction ALLOW pull for attraction ALLOW attracts with more force for greater attraction IGNORE just greater force (could be repulsion) IGNORE comparison of bond strength/energy to break bonds IGNORE comparisons of numbers of ions IGNORE responses in terms of packing 6

Question Answer Marks Guidance 2 (a) (i) (K c = ) [CO 2 ]2 [N 2 ] [CO] 2 [NO] 2 1 Square brackets required for ALL four concentrations (ii) dm 3 mol 1 1 ALLOW mol 1 dm 3 7

Question Answer Marks Guidance 2 (a) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 0.95 award 4 marks -------------------------------------------------------------------- Equilibrium amounts: n(co) = 0.46 0.20 = 0.26 mol n(co 2 ) = 0.2(0) mol n(n 2 ) = 0.1(0) mol K c calculation Must use calculated equilibrium amounts AND 0.25 0.20 2 0.10 3 1 (K c = ) = 0.95 (dm mol ) 0.26 2 0.25 2 4 ANNOTATIONS MUST BE USED IF there is an alternative answer, apply ECF by checking working for intermediate marks ---------------------------------------------------------------------------- APPLY ECF from incorrect starting n(co) By ECF, n(n 2 ) = n(co 2 )/2 For all parts, ALLOW numerical answers from 2 significant figures up to the calculator value Correct numerical answer with no working scores 4 marks ALLOW calculator value: 0.946745562 down to 0.95 (2SF), correctly rounded, e.g. 0.947 IGNORE units, even if incorrect --------------------------------------------------------------- Common errors 1.89 3 marks use of n(n 2 ) = 0.2(0) mol (K c = ) 0.202 0.20 3 1 0.26 2 = 1.893491124 (dm mol ) 2 0.25 1.29 3 marks 0.45 and 0.46 swapped over n(co) = 0.45 0.21 = 0.24 mol n(co 2 ) = 0.21 mol n(n 2 ) = 0.105 mol (K c = ) 0.212 0.105 3 1 0.24 2 = 1.28625 (dm mol ) 2 0.25 1.0243 marks 0.45 used twice n(co) = 0.45 0.20 = 0.25 mol n(co 2 ) = 0.2(0) mol n(n 2 ) = 0.1(0) mol (K c = ) 0.202 0.10 3 1 0.25 2 = 1.024 (dm mol ) 2 0.25 1.1853 marks 0.46 used twice n(co) = 0.46 0.21 = 0.25 mol n(co 2 ) = 0.21 mol n(n 2 ) = 0.105 mol (K c = ) 0.212 0.105 3 1 0.25 2 = 1.185408 (dm mol ) 2 0.25 8

Question Answer Marks Guidance 2 (a) (iv) Mark ECF from (iii) IF K c from (iii) < 1 equilibrium to left/towards reactants OR IF K c from (iii) > 1 equilibrium to right/towards products 1 First look at K c value for (iii) at bottom of cut ------------------------------------------------- ALLOW favours reverse reaction For correct K c value in (iii) of 0.95, ALSO ALLOW equilibrium position near to centre (b) (i) K c has decreased AND H is negative OR (forward) reaction is exothermic 1 Statement AND reason required for mark ALLOW for reason: reverse reaction is endothermic (ii) Effect of T and P on equilibrium (increased) temperature shifts equilibrium to left AND (increased) pressure shifts equilibrium to right AND fewer (gaseous) moles on right-hand side Reason ONLY required for pressure Temperature and H had been required in (i) ALLOW ratio of (gas) moles is 4:3 Overall effect on equilibrium Difficult to predict relative contributions of two opposing factors 2 Total 10 ALLOW opposing effects may not be the same size ALLOW effects could cancel each other out ALLOW effects oppose one another DO NOT ALLOW just it is difficult to predict equilibrium position (in question) For the 2nd mark, we are assessing the idea that we don t know which factor is dominant 9

Question Answer Marks Guidance 3 (a) (i) (K a =) [H+ ][CH 3 (CH 2 ) 2 COO ] ALLOW CH 3 CH 2 CH 2 COOH OR C 3 H 7 COOH in expression 1 [CH 3 (CH 2 ) 2 COOH] DO NOT ALLOW use of HA and A in this part. DO NOT ALLOW: [H + ][CH 3 (CH 2 ) 2 COO ] [CH 3 (CH 2 ) 2 COOH] = [H + ] 2 [CH 3 (CH 2 ) 2 COOH] : CON (ii) pk a = logk a = 4.82 1 ALLOW 4.82 up to calculator value of 4.821023053 DO NOT ALLOW 4.8 (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 2.71 award 3 marks -------------------------------------------------------------------- [H + ] = [K a ][CH 3 (CH 2 ) 2 COOH] OR 1.51 10 5 0.250 [H + ] = 1.94 x 10 3 (mol dm 3 ) IF alternative answer to more or fewer decimal places, check calculator value and working for 1st and 2nd marks ------------------------------------------------------------ ALLOW use of HA and A in this part Calculator: 1.942935923 x 10 3 ALLOW use of calculated K a value, either calculator value or rounded on script. ph must be to 2 decimal places ph = log[h + ] = 2.71 3 ALLOW ECF from incorrectly calculated [H + ] and ph ONLY when values for both K a AND [CH 3 CH 2 CH 2 COOH] have been used, i.e. 1.5 x 10 5 AND 0.250. e.g.: ph = 5.42 2 marks ph = 2.11 2 marks log(1.51 x 10 5 x 0.250) No 1.51 10 5 log( ) 0.250 1.51 10 5 ph = 4.22 1 mark log( ) No 0.250 DO NOT ALLOW just log(1.51 x 10 5 ) = 4.82 NO MARKS 10

Question Answer Marks Guidance 3 (b) (i) Mg + 2H + Mg 2+ + H 2 1 IGNORE state symbols ALLOW Mg + 2 CH 3 (CH 2 ) 2 COOH 2CH 3 (CH 2 ) 2 COO + Mg 2+ + H 2 DO NOT ALLOW on RHS: (CH 3 (CH 2 ) 2 COO ) 2Mg 2+ Ions must be shown separately (ii) CO 3 2 + 2H + H 2 O + CO 2 1 IGNORE state symbols ALLOW CO 3 2 + 2 CH 3 (CH 2 ) 2 COOH 2 CH 3 (CH 2 ) 2 COO + H 2 O + CO 2 ALLOW as product H 2 CO 3 (c) (i) CH 3 (CH 2 ) 2 COONa OR CH 3 (CH 2 ) 2 COO forms OR CH 3 (CH 2 ) 2 COOH + OH CH 3 (CH 2 ) 2 COO + H 2 O CH 3 (CH 2 ) 2 COOH is in excess OR acid is in excess OR some acid remains 2 ALLOW names throughout ALLOW sodium salt of butanoic acid ALLOW CH 3 (CH 2 ) 2 COOH + NaOH CH 3 (CH 2 ) 2 COONa + H 2 O DO NOT ALLOW just forms a salt/conjugate base i.e. identity of product is required 11

Question Answer Marks Guidance 3 (c) (ii) Moles (2 marks) ANNOTATIONS MUST BE USED amount CH 3 (CH 2 ) 2 COOH = 0.0100 (mol) -------------------------------------------------------------------------------- ALLOW HA and A throughout amount CH 3 (CH 2 ) 2 COO = 0.0025 (mol) 2 Concentration (1 mark) [CH 3 (CH 2 ) 2 COOH] = 0.100 mol dm 3 AND [CH 3 (CH 2 ) 2 COO ] = 0.025 mol dm 3 [H + ] and ph (2 marks) [H + ] = 1.51 10 5 0.100 0.025 = 6.04 x 10-5 (mol dm 3 ) 1 2 Mark by ECF throughout ONLY award final 2 marks via a correct ph calculation via K a [CH (CH ) COOH] 3 2 2 using data derived from that in the [CH 3 (CH 2 ) 2 COO ] question (i.e. not just made up values) ph = log 6.04 x 10-5 = 4.22 ph to 2 DP ALLOW alternative approach based on Henderson Hasselbalch equation for final 2 marks 0.025 ph = pk a + log OR pk a log 0.100 0.100 0.025 ph = 4.82 0.60 = 4.22 ALLOW logk a for pk a TAKE CARE with awarding marks for ph = 4.22 There is a mark for the concentration stage. If this has been omitted, the ratio for the last 2 marks will be 0.0100 and 0.0025. 4 marks max. Common errors ph = 5.42 As above for 4.22 but with acid/base ratio inverted. Award 4 OR 3 marks Award zero marks for: 4.12 from no working or random values ph value from K a square root approach (weak acid ph) ph value from K w /10 14 approach (strong base ph) Common errors ph = 4.12 use of initial concentrations: 0.250 and 0.050 given in question. Award last 3 marks for: 0.250/2 AND 0.050/2 = 0.125 AND 0.025 1.51 10 5 0.125 0.025 = 7.55 x 10-5 (mol dm 3 ) ph = log[h + ] = 4.12 Award last 2 marks for: 1.51 10 5 0.250 0.050 = 7.55 x 10-5 (mol dm 3 ) ph = log[h + ] = 4.12 ph = 5.52 As above for 4.12 but with acid/base ratio inverted. Award 2 OR 1 marks as outlined for 4.12 above 12

Question Answer Marks Guidance 3 (d) HCOOH + CH 3 (CH 2 ) 2 COOH State symbols NOT required HCOO + + CH 3 (CH 2 ) 2 COOH ALLOW 1 and 2 labels the other way around. 2 ALLOW just acid and base labels throughout if linked by lines so that it is clear what the acid-base pairs are acid 1 base 2 base 1 acid 2 CARE: Both + and charges are required for the products in the equilibrium DO NOT AWARD the 2nd mark from an equilibrium expression that omits either charge 2 Total 16 For 1st mark, DO NOT ALLOW COOH (i.e. H at end rather than start) but within 2nd mark ALLOW COOH by ECF IF proton transfer is wrong way around then ALLOW 2nd mark for idea of acid base pairs, i.e. HCOOH + CH 3 (CH 2 ) 2 COOH HCOOH 2 + + CH 3 (CH 2 ) 2 COO base 2 acid 1 acid 2 base 1 For H 2 COOH + shown with wrong proton transfer, DO NOT ALLOW an ECF mark for acid base pairs 13

Question Answer Marks Guidance 4 (a) (i) initial rates data: From Experiment 1 to Experiment 2 AND [NO 2 ] x 1.5, rate x 1.5 ANNOTATIONS MUST BE USED Quality of Written Communication: -------------------------------------------------------------------------------- Changes MUST be linked to Experiment numbers in writing (Could be described unambiguously) IGNORE annotations in the table -------------------------------------------------------------------------------- For 2nd condition, ALLOW when [NO 2 ] increases by half, rate increases by half 1st order with respect to NO 2 NOTE: Orders may be identified within a rate equation From Experiment 2 to Experiment 3 AND [O 3 ] is doubled, rate x 2 1st order with respect to O 3 rate equation and rate constant: rate = k[no 2 ] [O 3 ] rate k = [NO2 ][O 3 ] OR 4.80 10 8 0.00150 0.00250 = 0.0128 dm 3 mol 1 s 1 8 ALLOW: working from any of the Experiments : All give the same calculated answer 0.0128 subsumes previous rearrangement mark ALLOW: mol 1 dm 3 s 1 DO NOT ALLOW 0.013 over-rounding ----------------------------------------------------------------------------- [NO2 ALLOW ECF from inverted k expression: k = ][O ] 3 : rate k = 78.125 ALLOW 3 SF or more NOTE units must be from rate equation 14

Question Answer Marks Guidance 4 (a) (ii) step 1: NO 2 + O 3 LHS of step one NO 3 + O 2 State symbols NOT required For rest of equations, ALLOW other combinations that step 2: NO 2 + NO 3 N 2 O 5 together give the overall equation, rest of equations for step 1 AND step 2 2 e.g.: NO 5 NO 2 + NO 5 N 2 O 5 + O 2 CHECK that each equation is balanced CARE: Step 1 AND Step 2 must add up to give overall equation In Step 2, IGNORE extra species shown on both sides, e.g. NO 2 + NO 3 + O 2 N 2 O 5 + O 2 Step 2 can only gain a mark when Step 1 is correct e.g.: NO + 2O 2 NO + NO 2 + O 2 N 2 O 5 DO NOT ALLOW use of algebraic species, e.g. X (b) (i) 3 gaseous moles 2 gaseous moles Less randomness OR becomes more ordered 2 ALLOW products have fewer gaseous moles ORA ALLOW molecules instead of moles ALLOW fewer ways of distributing energy OR fewer degrees of freedom OR fewer ways to arrange (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 148 award 3 marks -------------------------------------------------------------------- G = H T S = 198 (298 x 168/1000) = 148 (kj mol 1 ) 3 IF there is an alternative answer, check calculator value and working for intermediate marks by ECF ------------------------------------------------------------ 2nd mark subsumes 1st mark for G = H T S ALLOW 148 to calculator value of 147.936 ALLOW for 2 marks: 49866 (kj mol 1 ): not converting S from J to kj (no 1000) 193.8 (kj mol 1 ) use of 25 instead of 298 15

Question Answer Marks Guidance 4 (b) (iii) CARE: responses involve changes of negative values -------------------------------------------------------------------------- ANNOTATIONS MUST BE USED Feasibility with increasing temperature Reaction becomes less feasible/not feasible AND G increases OR G becomes less negative OR G = 0 OR G > 0 OR G is positive OR G approaches zero ***IF a candidate makes a correct statement about the link between G and feasibility, IGNORE an incorrect H and T S relationship IF there is no G statement, then mark any H and T S relationship in line with the mark scheme --------------------- Effect on T S T S becomes more negative OR T S decreases OR T S increases OR magnitude of T S increases 2 -------------------------------------------------------------------------------- As alternative for not feasible ALLOW not spontaneous OR a comment that implies reaction does not take place ALLOW for G increases H = T S OR H > T S OR H T S is positive OR T S becomes more significant than H OR T S becomes the same as H OR T S becomes more negative than H NOTE Last statement will also score 2nd mark ------------------- DO NOT ALLOW T S increases --------------------------------------------------------------------------------- Total 17 ---------------------------------------------------------------------- APPROACH BASED ON TOTAL ENTROPY: Feasibility with increasing temperature Reaction becomes less feasible/not feasible AND S H/T OR S total decreases/ less positive OR S outweighs/ is less significant than H/T Effect on H/T H/T is less negative OR H/T increases OR H/T decreases OR magnitude of H/T decreases 16

Question Answer Marks Guidance 5 (a) (A transition element) has (at least) one ion with a partially filled d sub-shell/ d orbital ALLOW incomplete for partially filled DO NOT ALLOW d shell Fe AND 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 Fe(II) / Fe 2+ AND 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 Fe(III) / Fe 3+ AND 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4 ALLOW 4s before 3d, i.e. 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 IF candidate has used subscripts OR caps OR [Ar], DO NOT ALLOW when first seen but credit subsequently, i.e. 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3D 6 [Ar]4s 2 3d 6 For Fe 2+ and Fe 3+, ALLOW 4s 0 in electron configuration IGNORE electron configurations of elements other than Fe (b) EXAMPLES MUST REFER TO Cu 2+ FOR ALL MARKS PRECIPITATION Reagent NaOH(aq) OR KOH(aq) States not required ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- ALLOW NaOH in equation if reagent not given in description ALLOW a small amount of NH 3 /ammonia DO NOT ALLOW concentrated NH 3 DO NOT ALLOW just OH Transition metal product AND observation Cu(OH) 2 AND blue precipitate/solid Correct balanced equation Cu 2+ (aq) + 2OH (aq) Cu(OH) 2 (s) state symbols not required IF more than one example shown, mark example giving lower mark 3 ALLOW Cu(OH) 2 (H 2 O) 4 ALLOW any shade of blue ALLOW (s) as state symbol for ppt (may be in equation) ALLOW [Cu(H 2 O) 6 ] 2+ + 2OH Cu(OH) 2 (H 2 O) 4 + 2H 2 O For NH 3, also ALLOW: + [Cu(H 2 O) 6 ] 2+ + 2NH 3 Cu(OH) 2 (H 2 O) 4 + 2NH 4 ALLOW full equation, e.g. CuSO 4 + 2NaOH Cu(OH) 2 + Na 2 SO 4 CuCl 2 + 2NaOH Cu(OH) 2 + 2NaCl 17

Question Answer Marks Guidance 5 (b) LIGAND SUBSTITUTION 2 likely Reagent NH 3 (aq)/ammonia State not required IF more than one example shown, mark example giving lower mark ALLOW NH 3 in equation if reagent not given in description Transition metal product AND observation [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ AND deeper/darker blue (solution) Correct balanced equation [Cu(H 2 O) 6 ] 2+ + 4NH 3 [Cu(NH 3 ) 4 (H 2 O) 2 ] 2+ + 4H 2 O OR ------------------------------------------------------- Reagent Concentrated HCl OR (dilute) HCl(aq) OR NaCl(aq) State not required Transition metal product AND observation [ CuCl 4] 2 AND yellow (solution) Correct balanced equation [Cu(H O) ] 2+ + 4Cl [CuCl ] 2 + 6H O 2 6 4 2 3 DO NOT ALLOW precipitate ALLOW royal blue, ultramarine blue or any blue colour that is clearly darker than for [Cu(H 2 O) 6 ] 2+ ----------------------------------------------- ALLOW CuCl 4 2 i.e. no brackets ALLOW any shades of yellow, e.g. yellow green DO NOT ALLOW precipitate ALLOW other correct ligand substitutions using same principles for marking as in two examples given (c) (i) Pt oxidised from 0 +4 N reduced from +5 to +4 2 ALLOW 1 mark for Pt from 0 to +4 AND N from +5 to +4 i.e. oxidation and reduction not identified or wrong way round DO NOT ALLOW Pt is oxidised and N reduced with no evidence DO NOT ALLOW responses using other incorrect oxidation numbers (CON) 18

Question Answer Marks Guidance 5 (c) (ii) Pt + 6HCl + 4HNO 3 H 2 PtCl 6 + 4NO 2 + 4H 2 O 1st mark for ALL species correct and no extras: i.e: 2 Pt + HCl + HNO 3 H 2 PtCl 6 + NO 2 + H 2 O DO NOT ALLOW charge on Pt, e.g. Pt 2+ 2nd mark for correct balancing ALLOW correct multiples (d) Cl Cl Cl Pt Cl Cl Cl 2 OR Must contain 2 out wedges, 2 in wedges and 2 lines in plane of paper OR 4 lines, 1 out wedge and 1 in wedge For bond into paper, ALLOW: IGNORE charges on Pt and Cl for this mark The 2 marks for charge AND bond angle are ONLY available from a diagram showing Pt bonded to 6 Cl ONLY 3-D Shape 1 mark Correct 3-D diagram of Pt surrounded by 6Cl ONLY Bond angle 1 mark bond angle of 90º on diagram or stated 3 ALLOW ONLY if diagram has Pt surrounded by 6Cl ONLY BUT 3-D shape may not be correct DO NOT ALLOW if ANY charges shown on Pt or Cl within brackets Charge 1 mark 2 charge shown outside of brackets 19

Question Answer Marks Guidance 5 (e) (i) Donates two electron pairs to a metal (ion) ALLOW lone pairs for electron pairs forms two coordinate bonds 2 ALLOW dative (covalent) bond for coordinate bond ALLOW 1 mark for a full definition of a ligand (without reference to 2: i.e. Donates an electron pair to a metal (ion) forming a coordinate bond (ii) NH 2 O O ALLOW displayed formulae charges essential in (COO ) 2 structure DO NOT ALLOW H 2 N NH 2 O O 2 Total 21 20

Question Answer Marks Guidance 6 (a) (i) complete circuit with voltmeter and salt bridge linking two half-cells Salt bridge MUST be labelled Pt electrode in Fe 3+ /Fe 2+ half-cell with same concentrations Cr electrode in 1 mol dm 3 Cr 3+ half-cell 3 ALLOW Fe 2+ and Fe 3+ with concentrations of 1 mol dm 3 ALLOW 1 M but DO NOT ALLOW 1 mol (ii) Cr + 3Fe 3+ Cr 3+ + 3Fe 2+ 1 ALLOW sign (iii) 1.51 V 1 IGNORE sign DO NOT ALLOW if e shown uncancelled on both sides, e.g. Cr + 3Fe 3+ + 3e Cr 3+ + 3Fe 2+ + 3e (b) Cr 2 O 7 2 AND H + 1 ALLOW acidified dichromate (c) Cr 2 O 2 7 (aq) + 8H + (aq) + 3HCOOH(aq) 2Cr 3+ (aq) + 7H 2 O(l) + 3CO 2 (l) State symbols not required 2 1st mark for ALL species correct and no extras: Cr 2 O 7 2, H +, HCOOH, Cr 3+, H 2 O AND CO 2 NOTE: H + may be shown on both sides ALLOW sign (d) (i) E o for chromium (redox system) is more negative/lower/less (than copper redox system) ORA 2nd mark for correct balancing with H + cancelled down ALLOW E cell is +1.08 V (sign required) chromium system shifts to the left / Cr(s) Cr 3+ (aq) + 3e AND copper system shifts to the right / Cu 2+ (aq) + 2e Cu(s) 2 ALLOW Cr loses electrons more readily/more easily oxidised OR Cr is a stronger reducing agent OR Cu loses electrons less readily OR Cu is a weaker reducing agent 21

Question Answer Marks Guidance 6 (d) (ii) Cr reacts with H + ions/acid to form H 2 gas 1 ALLOW equation: 2Cr + 6H + 2Cr 3+ + 3H 2 (ALLOW multiples) DO NOT ALLOW just hydrogen forms, i.e. Cr, H + /acid AND H 2 must all be included for the mark (e) (i) 1.45 V 1 IGNORE sign (ii) 2 marks,, for two points from the following list: 2 1. Methanoic acid is a liquid AND easier to store/transport OR hydrogen is a gas AND harder to store/transport OR hydrogen as a liquid is stored under pressure 2. Hydrogen is explosive/more flammable 3. HCOOH gives a greater cell potential/voltage ASSUME it refers to HCOOH DO NOT ALLOW 'produces no CO 2 ' IGNORE comments about biomass and renewable HCOOH and H 2 are both manufactured from natural gas 4. HCOOH has more public/political acceptance than hydrogen as a fuel Total 14 22

Question Answer Marks Guidance ALLOW e : 7 (a) MnO 2 + 4OH MnO 4 2 + 2H 2 O + 2e i.e. sign not required 3H 2 O + ClO 3 + 6e 6OH + Cl 2 (b) Role of CO 2 CO 2 reacts with H 2 O forming an acid OR carbonic acid/h 2 CO 3 forms OR CO 2 is acidic Equation involving OH H 2 CO 3 + OH H 2 O + HCO 3 OR H 2 CO 3 + 2OH 2H 2 O + CO 3 2 OR CO 2 + OH CO 3 2 + H + OR CO 2 + OH HCO 3 OR CO 2 + 2OH CO 3 2 + H 2 O OR H + + OH H 2 O ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- ALLOW equation: CO 2 + H 2 O H 2 CO 3 OR CO 2 + H 2 O H + + HCO 3 OR CO 2 + H 2 O 2H + + CO 3 2 Effect on equilibrium with reason equilibrium shifts to right AND to restore OH 3 ALLOW for restores OH the following: makes more OH, OH has been used up DO NOT ALLOW just equilibrium shifts to right 23

Question Answer Marks Guidance 7 (c) FOLLOW through stages to mark -------------------------------------------------------------------------- Moles in titration n(kmno 4 ) = 0.0200 x 26.2 1000 = 5.24 x 10 4 mol ANNOTATIONS MUST BE USED AT LEAST 3 SF for each step -------------------------------------------------------------------------------- n(so 3 2 ) = 1.31 x 10 3 mol Scaling n(so 3 2 - ) in original 100 cm 3 = 4 x 1.31 x 10 3 = 5.24 x 10 3 mol ECF 2.5 x answer above ECF 4 x answer above Mass Mass of Na 2 SO 3 in sample = 126.1 x 5.24 x 10 3 g = 0.660764 g ECF 126.1 x answer above ALLOW 0.661 g up to calculator value Percentage % Na 2 SO 3 = 0.660764 0.720 100 = 91.8% 5 ALLOW alternative approach based on theoretical content of Na 2 SO 3 for last 2 marks Theoretical amount, in moles, of Na 2 SO 3 in sample 0.720 n(na 2 SO 3 ) = 126.1 = 5.71 x 10 3 mol Percentage % Na 2 SO 3 = 5.24 10 3 100 = 91.8% 3 5.71 10 calculated mass above ECF 100 0.720 ALLOW 91.8% (1 DP) up to calculator value of 91.77277778 i.e. DO NOT ALLOW 92% COMMON ERRORS: 36.8(1)% 4 marks no 2.5 factor 22.9(4)% 4 marks no scaling by 4 9.18% 3 marks no 2.5 and no x 4 Watch for random ECF %s for % from incorrect M(Na 2 SO 3 ), e.g. use of M(SO 3 2 ) = 80.1 giving 58.3% Total 10 24

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